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Precalculus : Functions

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Example Questions

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Example Question #2 : Triangles

Let ABC be a right triangle with sides = 3 inches, = 4 inches, and = 5 inches. In degrees, what is the $\sin \Theta$ where $\Theta$ is the angle opposite of side ?

Possible Answers:

Correct answer:

Explanation:

3-4-5_triangle

We are looking for $\sin(\theta )$ . Remember the definition of sin in a right triangle is the length of the opposite side divided by the length of the hypotenuse.

So therefore, without figuring out $\Theta$ we can find

$\sin \Theta =\frac{3}{5}=.6$

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Example Question #11 : Functions

Simplify the function below:

$f(x)=cos^2(\frac{1}{2}cos^{-1}x)$

Possible Answers:

$\frac{x}{2}$

cosx

$\frac{1-x}{2}$

$\frac{1+x}{2}$

Correct answer:

$\frac{1+x}{2}$

Explanation:

We need to use the following formulas:

a) $cos^2x=\frac{1+cos2x}{2}$

and

b) $cos(cos^{-1}x)=x$

We can simplify f(x) as follows:

$f(x)=cos^2(\frac{1}{2}cos^{-1}x)=\frac{1+cos(cos^{-1}x)}{2}=\frac{1+x}{2}$

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Example Question #11 : Relations And Functions

Simplify

$\cot(x)\times{\sin(x) }$

Possible Answers:

$\arcsin(x)$

$\csc(x)$

$\arcsin(\pi )$

$\cos(x)$

$\tan(x)$

Correct answer:

$\cos(x)$

Explanation:

$\cot(x) = \cos(x)/\sin(x)$ . Thus: $[\cos(x)/\sin(x)]\times\sin(x)=\cos(x)$

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Example Question #1 : Trigonometric Identities

Simplify

$\frac{cot (x)\cdot tan(x)}{cos(x)}$

Possible Answers:

$\cot(x)/sec(x)$

$\csc(x)$

$\cos(x)\times\tan(x)$

$\sec(x)$

$\cos(x)$

Correct answer:

$\sec(x)$

Explanation:

$\cot(x)\times\tan(x) = 1$

and

$1/(\cos(x)) = \sec(x)$ .

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Example Question #5 : Solving Trigonometric Equations

If $sinx-cosx=\frac{1}{3}$ , find the value of the function $f(x)=sin^{3}x+cos^{3}x$ .

Possible Answers:

$-\frac{13}{27}$

$\frac{11}{27}$

$-\frac{11}{27}$

$\frac{13}{27}$

Correct answer:

$\frac{13}{27}$

Explanation:

We can write:

$sinx-cosx=\frac{1}{3}$

$\Rightarrow (sinx-cosx)^{2}=\frac{1}{9}$

$\Rightarrow sin^2x+cos^2x-2sinxcosx=\frac{1}{9}$

$\Rightarrow 2sinxcosx=1-\frac{1}{9}$

$\Rightarrow sinxcosx=\frac{4}{9}$

We can also write:

(sinx-cosx)^3=sin^3x+cos^3x+3sinxcos^2x-3cosxsin^2x

$\Rightarrow sin^3x+cos^3x=(sinx-cosx)^3+3sinxcosx(sinx-cosx)$

Now we can substitute the values of (sinx - cosx) and (sinx * cosx) in the obtained function:

$\Rightarrow sin^3x+cos^3x=(\frac{1}{3})^3+3*\frac{4}{9}*\frac{1}{3}=\frac{1}{27}+\frac{4}{9}=\frac{13}{27}$

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Example Question #11 : Relations And Functions

Identifying a function.

Which of the following is a function.

Possible Answers:

$g(x)=\pm \sqrt{x^2-9}$

h(x)=-3x^4-2x+3sinx

$x=\left | y \right |$

(x-4)^2+y^2=25

x=2y^2-14

Correct answer:

h(x)=-3x^4-2x+3sinx

Explanation:

The only relation listed that doesn't map more than one dependent variable value for some independent variable value is h(x). Another way of saying this is that each value in the domain of h corresponds to a distinct value in its range. h(x) is the only one-to-one relation, and so is the only option that is a function.

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Example Question #12 : Relations And Functions

Evaluating a function.

Evaluate f(-3) , when

$f(x)=\frac{3x^2-4x+1}{x+7}$ .

Possible Answers:

f(-3)=4

$f(-3)=1\tfrac{3}{5}$

f(-3)=1.6

f(-3)=10

$f(-3)=-\frac{1}{2}$

Correct answer:

f(-3)=10

Explanation:

Plug in -3 for x in the function rule:

$\frac{3(-3)^2-4(-3)+1}{-3+7}=\frac{3\cdot 9+12+1}{4}=\frac{27+12+1}{4}=\frac{40}{4}=10$ .

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Example Question #43 : Pre Calculus

Finding inverse functions.

Given,

$h(x)=\frac{5x+2}{x-3}$

find

$h^{-1}(x)$ .

Possible Answers:

$h^{-1}(x)=\frac{5x+2}{x-5}$

$h^{-1}(x)=\frac{-2}{3}$

$h^{-1}(x)=\frac{3y+2}{y-5}$

$h^{-1}(x)=\frac{5x+2}{x-3}$

$h^{-1}(x)=\frac{3x+2}{x-5}$

Correct answer:

$h^{-1}(x)=\frac{3x+2}{x-5}$

Explanation:

To find an inversefunction find the input in terms of the output. In other words, solve the function for the dependent variable.

First, set h(x) = y. Now solve for y.

$y=\frac{5x+2}{x+3}$

Multiply the denominator on both sides

y(x-3)=5x+2

Distribute y

xy-3y=5x+2

Rearrange to get only x terms on one side.

xy-5x=3y+2

Factor out x on the left side.

x(y-5)=3y+2

Divide by (y-5) on both sides to get x by itself.

$x=\frac{3y+2}{y-5}$

The final step is to recognize that the independent variable is now the dependent variable. To put this into proper notation switch x and y. The inverse function is

$h^{-1}(x)=\frac{3x+2}{x-5}$ .

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Example Question #51 : Pre Calculus

Finding an inverse function.

Given

f(x)=2x+1

find

f^-1(x) .

Possible Answers:

$f^{-1}(x)={x+2}$

$f^{-1}(x)=2x-\frac{1}2{}$

$f^{-1}(x)=\frac{1}{2}x+1$

$f^{-1}(x)=x$

$f^{-1}(x)=\frac{1}{2}x-\frac{1}{2}$ .

Correct answer:

$f^{-1}(x)=\frac{1}{2}x-\frac{1}{2}$ .

Explanation:

Finding an inverse function is essentially finding the output in terms of the input. To do this switch the independent and dependent variables. First set

f(x)=y .

Then

y=2x+1

Now, to switch the role of the variables solve for x in terms of y.

y-1 = 2x

2x=y-1

$x=\frac{1}2{}y-\frac{1}{2}$

So, y is now the independent variable, and x is the dependent variable. The final step is to write the inverse function in proper notation.

Set $x=f^{-1}(x)$ , and y=x .

This gives

$f^{-1}(x)=\frac{1}{2}x-\frac{1}{2}$ .

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Example Question #11 : Functions

Which function below correctly translates f(x) four units to the right and two units down?

Possible Answers:

f(x+4)+2

f(x+4)-2

f(x-4)-2

f(x-4)+2

Correct answer:

f(x-4)-2

Explanation:

The translation f(x+h) moves the graph of f(x) units to the left if h>0 . To see this, compare the graphs of y=x and y=x+1 . A translation to the right must be in the form f(x-h) if h>0 .

A vertical translation is up if k>0 . We want to move the function down, so our translation subtracts two from f(x-h) .

Combining these, we get f(x-4)-2 .

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Precalculus : Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #2 : Triangles

Example Question #11 : Functions

Example Question #11 : Relations And Functions

Example Question #1 : Trigonometric Identities

Example Question #5 : Solving Trigonometric Equations

Example Question #11 : Relations And Functions

Example Question #12 : Relations And Functions

Example Question #43 : Pre Calculus

Example Question #51 : Pre Calculus

Example Question #11 : Functions

All Precalculus Resources

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